18. \( \left(3-\frac{x}{3}\right)\left(3+\frac{x}{3}\right)\left(3-\frac{x}{3}\right)\left(3+\frac{x}{3}\right) \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Beyond the Answer
To simplify \( \left(3-\frac{x}{3}\right)\left(3+\frac{x}{3}\right)\left(3-\frac{x}{3}\right)\left(3+\frac{x}{3}\right) \), we can first recognize that \( \left(3 - \frac{x}{3}\right)\left(3 + \frac{x}{3}\right) \) resembles the difference of squares pattern. Let \( a = 3 \) and \( b = \frac{x}{3} \). So we can rewrite the expression: \[ (3 - \frac{x}{3})(3 + \frac{x}{3}) = 3^2 - \left(\frac{x}{3}\right)^2 = 9 - \frac{x^2}{9} = \frac{81 - x^2}{9} \] Now, we notice that there are two identical pairs of these products: \[ \left((3 - \frac{x}{3})(3 + \frac{x}{3})\right)^2 = \left(\frac{81 - x^2}{9}\right)^2 = \frac{(81 - x^2)^2}{81} \] Putting it all together, we find: \[ \left(3 - \frac{x}{3}\right)\left(3 + \frac{x}{3}\right)\left(3 - \frac{x}{3}\right)\left(3 + \frac{x}{3}\right) = \frac{(81 - x^2)^2}{81} \] So, the final answer is: \[ \frac{(81 - x^2)^2}{81} \]